Correct formula to use for potential difference?

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SUMMARY

The correct formula for calculating potential difference (ΔV) in an electric field is ΔV = V2 - V1 = -E*Δs, where E represents the electric field strength and Δs is the displacement vector pointing from point 1 to point 2. The negative sign indicates that the electric field direction opposes the direction of the potential difference. This relationship is crucial when analyzing scenarios such as capacitor plates, where the potential difference is positive when moving from a negatively charged plate to a positively charged plate. The formula E = -ΔV/Δs further clarifies the relationship between electric field and potential difference.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the concept of potential difference
  • Knowledge of vector notation in physics
  • Basic principles of electrostatics, particularly in capacitors
NEXT STEPS
  • Study the derivation of the relationship between electric field and potential difference
  • Explore the concept of electric field lines and their implications on potential difference
  • Learn about the behavior of capacitors in circuits and their voltage characteristics
  • Investigate the implications of charge interactions in electric fields
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Physics students, electrical engineers, educators, and anyone seeking a deeper understanding of electric fields and potential difference calculations.

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My book can't stick to one formula EVER.
In Q7 They start off using: E = −ΔV Δs (equation 20-4)

Then magically go to: ΔV = EΔs

Where is the negative sign because it never states to find the magnitude of the potential difference

http://img198.imageshack.us/img198/6914/msbn.jpg




I don't see how ΔV is directionless because in problem 3 it takes on the negative sign


http://img90.imageshack.us/img90/6047/pa9.gif
 
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You copied one of the equations wrong: you wrote " E = - ΔV Δs. "

Here's the story: (vectors in bold)

the correct formula is V2 - V1 = ΔV = - Es where Δs is defined positive pointing from point 1 to point 2
and E is the force on a 1 Coulomb charge anywhere between points 1 and 2.

BUT: the 1 coulomb charge is assumed to not affect the given E field. So, more correctly,
E = limit as q → 0 of (force on a positive charge q)/q.

Example: capacitor plates with positive voltage on plate 2 and negative voltage on plate 1.
s points from plate 1 to plate 2.
ΔV = V2 - V1 = -Es > 0.
where Δs is positive since we're going from 1 to 2.
So E = - ΔV/Δs which would be negative. The force on a small positive test charge q is always qE so the force is negative and points in the direction opposite to Δs.
 

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